Relevant For...

The concepts of bijection, injection, and surjection are fundamental properties of functions. They allow for comparisons between cardinalities of sets, and are useful in proofs that two finite sets have the same size.

The function \( f\colon \{ \text{German football players dressed for the 2014 World Cup final}\} \to {\mathbb N} \) defined by \(f(A) = \text{the jersey number of } A\) is injective; no two players were allowed to wear the same number.

The existence of an injective function gives information about the relative sizes of its domain and range:

Let f be a One-one (Injective) function, with domain \(D_{f} = \{x,y,z\} \)and Range \(\{1,2,3\}\). It is given that only one of the following \(3\) statement is true and remaining statements are false.

Surjective

Let \(f \colon X\to Y\) be a function. Then \(f\) is surjective if every element of \(Y\) is the image of at least one element of \(X.\) That is, \( \text{image}(f) = Y.\)

Symbolically,

\[
\forall y \in Y, \exists x \in X \text{ such that } f(x) = y.
\]

A synonym for "surjective" is "onto."

The function \( f\colon {\mathbb Z} \to {\mathbb Z}\) defined by \( f(n) = 2n\) is not surjective: there is no integer \( n\) such that \( f(n)=3,\) because \( 2n=3\) has no solutions in \( \mathbb Z.\) So \( 3\) is not in the image of \( f.\)

Bijective

A function is bijective for two sets if every element of one set is paired with only one element of a second set, and each element of the second set is paired with only one element of the first set. This means that all elements are paired and paired once.

Let \(f \colon X \to Y \) be a function. Then \(f\) is bijective if it is injective and surjective; that is, every element \( y \in Y\) is the image of exactly one element \( x \in X.\)

The function \( f\colon \{ \text{months of the year}\} \to \{1,2,3,4,5,6,7,8,9,10,11,12\} \) defined by \(f(M) = \text{ the number } n \text{ such that } M \text{ is the } n\text{th month}\) is a bijection.

The following alternate characterization of bijections is often useful in proofs:

Suppose \( X \) is nonempty. Then \( f \colon X \to Y \) is a bijection if and only if there is a function \( g\colon Y \to X \) such that \( g \circ f \) is the identity on \( X \) and \( f\circ g\) is the identity on \( Y;\) that is, \(g(f(x))=x\) and \( f(g(y))=y \) for all \(x\in X, y \in Y.\) When this happens, the function \( g \) is called the inverse function of \( f \) and is also a bijection.

Show that the function \( f\colon {\mathbb R} \to {\mathbb R} \) defined by \( f(x)=x^3\) is a bijection.

Rather than showing \(f\) is injective and surjective, it is easier to define \( g\colon {\mathbb R} \to {\mathbb R}\) by \(g(x) = x^{1/3} \) and to show that \( g\) is the inverse of \( f.\) This follows from the identities \( (x^3)^{1/3} = (x^{1/3})^3 = x.\) (Followup question: the same proof does not work for \( f(x) = x^2.\) Why not?)